Question

A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0 west of north. How much work must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0° south of east?

Answer 1

Answer:

Explanation:

mass of crate, m = 30 kg

initial velocity, u = 3.9 m/s at an angle 37° west of north

final velocity, v = 5.62 m/s at an angle 63° south of east

By teh work energy theorem, the work done is equal to the change in kinetic energy of the body.

$W = \frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}$

W = 0.5 x 30 x (5.62² - 3.9²)

W = 15 x 16.37

W = 245.6 J

Answer 2

Work done is the same as energy. The much work done on the crate to change its velocity to 5.62 m/s in a direction 63.0° south of east is 1350 Joule approximately.

Given that a 30 kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0 west of north.

The resultant velocity must be calculated first by using cosine formula.

where angle ∅ = 37 + 90 + 63 = 190°

$V^{2}$ = $3.9^{2} + 5.62^{2}$ - 2(3.9)(5.62) cos 190

$V^{2}$ = 46.7944 - 43.836 cos 190

$V^{2}$ = 46.7944 + 43.170

$V^{2}$ = 89.9644

Resultant velocity V = $\sqrt{89.96}$

V = 9.48 m/s

To calculate the much work that must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0° south of east, we will use the kinetic energy formula because work done is the same as energy.

W = 1/2m$V^{2}$

Where

m = 30 kg and V = resultant velocity = 9.48 m/s

W = 0.5 x 30 x 89.96

W = 1349.4 Joule

Therefore, the much work done on the crate to change its velocity to 5.62 m/s in a direction 63.0° south of east is 1350 Joule approximately.

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